We all know that if we Highest Common Factor (or Greatest Common Divisor {As you prefer}) and the Least Common Multiple two numbers , then ALL WE KNOW is ->
H.C.F $/cdot$ L.C.M $=$ Product of 2 numbers {Well,as pef my knowledge , this is only applicable for 2 numbers}
Is there any kind of proof of this famous relation ??
The simplest proof is probably the following. Let $$ a = \prod_{i=1}^{n} p_{i}^{e_{i}},\qquad b = \prod_{i=1}^{n} p_{i}^{f_{i}}, $$ with $p_{i}$ distinct primes, and $e_{i}, f_{i} \ge 0$. Then $$ \gcd(a, b) = \prod_{i=1}^{n} p_{i}^{\min(e_{i}, f_{i})},\qquad \operatorname{lcm}(a, b) = \prod_{i=1}^{n} p_{i}^{\max(e_{i}, f_{i})}. $$ Now note that $$\min(e_{i}, f_{i}) + \max(e_{i}, f_{i})= e_{i} + f_{i}.$$
Suppose $d$ is the gcd of $a$ and $b$. We wish to show that $c=\frac{ab}{d}$ is the least common multiple.
First, note that $d\vert b$, so $\frac{b}{d}$ is an integer, and so $c = a\frac{b}{d}$ is an integer multiple of $a$. Thus, $a\vert c$. Similarly, $b\vert c$.
Now suppose $a\vert c'$ and $b\vert c'$ for some $c'$. We want to show that $c\vert c'$. Write $c'=ma=nb$. Further, write $a=pd, b=qd$. Then $c'=mpd=nqd$. Thus, $mp=nq$. Now, $q$ and $p$ share no common factors; if they did, $d$ wouldn't be the greatest common divisor. Thus, it follows that $q\vert m$, to $m = rq$. Then we have $$\frac{c'}{c} = \frac{mpd}{ab/d} = \frac{mpd^2}{pd\cdot qd} = \frac{m}{q} = r$$ And so $c\vert c'$. Thus, $c$ is the least common multiple of $a$ and $b$.
